java - 将递归结果添加到 ArrayList

Question

我试图找到一个单词的所有排列并将其添加到数组列表中并返回数组列表。但是，我相信我的递归是正确的，但是将结果添加到 ArrayList 时存在问题。这就是我到目前为止所拥有的。我传递的参数是“eat”和“”，返回的是3次“tea”

public static ArrayList<String> permutations(String word, String beginning)
{
    int l = word.length();
    ArrayList<String> temp = new ArrayList<String>();

    if(l == 0)
        temp.add(beginning + word);
    else
    {
        char c = word.charAt(l-1);
        String blah = (beginning + c);
        word = word.substring(0, l-1);
        for(int i = 0; i < l; i++)
        {
            permutations(word, blah);
            temp.add(blah + word);
        }

    }
    return temp;
}

Answer 1

可能我对你的方法没有正确的想法来找到一个简单的解决方案，当我开始工作时，我最终得到了这个。我希望这不会有太大的偏差，并且仍然有帮助。输出为:

[tea, tae, eta, eat, ate, aet]

import java.util.ArrayList;

public class Perm {
    public static void main(String[] args) {
        ArrayList<String> perms = new ArrayList<String>();
        permutations("tea", perms);
        System.out.println(perms);
    }

    public static ArrayList<String> permutations(String word, ArrayList<String> perms)
    {
        int l = word.length();

        // If the word has only a single character, there is only
        // one permutation -- itself. So we add it to the list and return
        if (l == 1) {
            perms.add(word);
            return perms;
        }

        // The word has more than one character.
        // For each character in the word, make it the "beginning"
        // and prepend it to all the permutations of the remaining
        // characters found by calling this method recursively

        for (int i=0; i<word.length(); ++i) {
            char beginning = word.charAt(i);

            // Create the remaining characters from everything before
            // and everything after (but not including) the beginning char
            String blah = word.substring(0,i)+word.substring(i+1);

            // Get all the permutations of the remaining characters
            // by calling recursively
            ArrayList<String> tempArray = new ArrayList<String>();
            permutations(blah, tempArray);

            // Prepend the beginning character to each permutation and
            // add to the list
            for (String s : tempArray) {
                perms.add(beginning + s);
            }
        }

        return perms;
    }
}